Question: $\int (\sqrt[6]{x})^3\,dx=$ $+C$
Answer: At first it might seem as if we can't apply any rule we've learned to find the indefinite integral of a radical function. However, remember that any radical can be rewritten as a rational power. $\int (\sqrt[6]{x})^3\,dx=\int x^{^{\frac36}}\,dx$ Now we can integrate using the reverse power rule: $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ $\begin{aligned} \int (\sqrt[6]{x})^3\,dx&=\int x^{^{\frac{3}{6}}}\,dx \\\\ &=\dfrac{x^{^{\frac{3}{6}+1}}}{\dfrac{3}{6}+1}+C \\\\ &=\dfrac{2}{3} x^{^{\frac{3}{2}}}+C \end{aligned}$ In conclusion, $\int (\sqrt[6]{x})^3\,dx=\dfrac{2}{3} x^{^{\frac{3}{2}}}+C$